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As per analysis for previous years, it has been observed that students preparing for NEET find Physics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Biology due to their medical background. Furthermore, sections such as Physics are dominantly based on theories, laws, numerical in comparison to a section of Biology which is more of fact-based, life sciences, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. .

Q1. We have seen that a gamma-ray does of 3 Gy is lethal to half the people exposed to it. If the equivalent energy were absorbed as heat, what rise in body temperature would result:
•  300μK
•  700μK
•  455μK
•  390μK
Solution
We can relate an absorbed energy Q and the resulting temperature increase ∆T with relation Q=cm∆T. In that equation, m is the mass of the material absorbing the energy and c is the specific heat of the material. An absorbed dose of 3 Gy corresponds to an absorbed energy per unit mass of 3 J/kg. Let us assume that c the specific heat of human body, is the same as that of water, 4180 J/kg K. Then we find that ∆T=(Q/m)/c=3/4180=7.2×10-4 K=700μK Obviously the damage done by ionizing radiation has nothing to do with thermal heating. The harmful effects arise because the radiation damages DNA and thus interferes with the normal functioning of tissues in which it is absorbed

Q2. A wall is made up of two layers A and B. The thickness of the two layers is the same, but materials are different. The thermal conductivity of A is double than that of B. In thermal equilibrium the temperature difference between the two ends is 36℃. Then the difference of temperature at the two surfaces of A will be
•  6℃
•  12℃
•  18℃
•  24℃
Solution
Suppose thickness of each wall is x then (Q/t)combination=(Q/t)A⇒(KS A(θ12))/2x=(2KA(θ1-θ))/x ∵K_S=(2×2K×K)/((2K+K))=4/3 K and (θ12 )=36° ⇒(4/3 KA×36)/2x=(2KA(θ1-θ))/x Hence temperature difference across will Ais (θ1-θ)=12℃

Q3. The thermal conductivity of a rod is 2. What is its thermal resistivity??
•  0.5
•  1
•  0.25
•  2
Solution
Thermal resistivity = 1/(Thermal conductivity) = 1/2 = 0.5

Q4. Newton’s law of cooling is a special case of
•  Stefan’s law
•  Kirchhoff’s law
•  Wien’s law
•  Planck’s law
Solution
For small difference of temperature, it is the special case of Stefan’s law

Q5.  The spectral energy distribution of a star is maximum at twice temperature as that of sun. the total energy radiated by star is.
•  Twice as that of the sun
•  Same as that of the sun
•  Sixteen times as that of the sun
•  One-sixteenth of the sun
Solution
According to Stefan’s law E=σT4 Where σ is Stefan’s constant. Given, T=2Ts ∴E'=σ(2Ts )4=16σTs4=16Es Hence, total energy radiated by star is sixteen times as that of the sun.

Q6. A particular star (assuming it as a black body) has a surface temperature of about 5×104 K. The wavelength in nanometers at which its radiation becomes maximum is (b=0.0029 mK)
•  48(
•  58
•  60
•  70
Solution
According to Wien’s displacement law λm T=b orλm=b/T=0.0029/(5×104 )=58×10-9 m=58nm

Q7. On a cold morning, a metal surface will feel colder to touch than a wooden surface because
•  Metal has high specific heat
•  Metal has high thermal conductivity
•  Metal has low specific heat
•  Metal has low thermal conductivity
Solution
Heat passes quickly from the body into the metal which leads to a cold feeling

Q8. In Searle’s method for finding conductivity of metals, the temperature gradient along the bar
•  Is greater nearer the hot end
•  Is greater nearer to the cold end
•  Is the same at all points along the bar
•  Increases as we go from hot end to cold end
Solution
Is the same at all points along the bar

Q9. In which process, the rate of transfer of heat is maximum
•  Conduction
•  Convection
•  In all these, heat is transferred with the same velocity
Solution
Radiation is the fastest mode of heat transfer

Q10. The temperature at which the vapour pressure of a liquid becomes equals to the external (atmospheric) pressure is its:
•  Melting point
•  Sublimation point
•  Critical temperature
• Boiling point
Solution
At boiling point, vapour pressure becomes equal to the external pressure ## Want to know more

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BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: Thermal Physics Quiz 16
Thermal Physics Quiz 16
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